CHAPTER 6 BLM ANSWER KEY

BLM 6-2: Kinetic Energy/Skill Builder

Answers

1. The values for the kinetic energy for each mass are: 1.0 kg, 0.5 J; 2.0 kg, 1.0 J; 3.0 kg, 1.5 J; 4.0 kg, 2.0 J; 5.0 kg, 2.5 J; 6.0 kg, 3.0 J; 7.0 kg, 3.5 J; 8.0 kg, 4.0 J; 9.0 kg, 4.5 J; 10.0 kg, 5.0 J.

2.

3. This graph has a straight line of best fit with a positive slope (sloping upward to the right).

4. Based on this shape, there is a direct relationship between the mass of an object and its kinetic energy.

5. The values for the kinetic energy for each velocity are: 1.0 m/s, 0.5 J; 2.0 m/s, 2.0 J; 3.0 m/s, 4.5 J; 4.0 m/s, 8.0 J; 5.0 m/s, 12.5 J; 6.0 m/s, 18.0 J; 7.0 m/s, 24.5 J; 8.0 m/s, 32.0 J; 9.0 m/s, 40.5 J; 10.0 m/s, 50.0 J.

6.

7. This graph shows a curved shape, which is a parabola.

8. Based on this shape, the kinetic energy depends on the square of the velocity. For example, doubling the velocity will quadruple the kinetic energy.

BLM 6-3: Work and Energy/Problem Solving

Answer

The gravitational potential energy of the student + toboggan while at the top of the hill is:

Eg = mgh = (60 kg) (9.80 / m / ) (25 m) = 14 700 J
s2

Friction is negligible since the snow is very slippery. Thus mechanical energy will be conserved. This means that

Eg (at top) = Ek (at bottom).

Ek = 14 700 J

At the start of the pavement section, Ek1 = 14 700 J , and at the end (when stopped), Ek2 = 0 J . From the work-kinetic energy theorem, the work done by friction to stop the toboggan equal the change in its kinetic energy.

W = Ek2 – Ek1 = –14 700 J

F||d = –14 700 J

F||(16 m) = –14 700 J

F|| = / –14 700 J
16 m
F|| = / –919N

(The force is negative, since it is exerted backward on the toboggan to stop it.)

BLM 6-4: Chapter 6 Test/Assessment

Answers

1. W = F||d = (10.0 N) (0.50 m) = 5.0J

2. W = mgh = (110 kg) (9.80 / m / ) (2.8 m)
s2
= 3018.4 J = 3.0 ´ 103 J

3. W = Fd cos q = (160 N) (15 m) (cos 40°)
= 1838.5 J = 1.8 ´ 103 J

4. Ek = mv2 = (1000 kg) (25 / m / )2
s
= 312 500 J = 3 ´ 105 J

5. W = Ek = Ek2 – Ek1

W = / 1 / mv22 – 0
2
50 J = / 1 / (30 kg) (v22)
2

v22 = (50 J) ÷ (3.0 kg)

v2 = 3.3 m/s

CHAPTER 6 BLM ANSWER KEY

6. Eg = mgh
= (65 kg) (9.80 / m / ) (10.0 m)
s2
= 6370J = 6.4 ´ 103 J
7. Eg (top) = Ek (bottom)
6370 J = / 1 / mv2
2

v2 = (6370) ÷ (65 kg)

v = 14 m/s

8. The spring was compressed approximately 0.76 m.

Let the maximum compression of the spring be x. Consider x to be in the vertical direction.

Use the quadratic equation to solve for x.

Since the mass has gone below the reference level
of x = 0, the value of x must be negative. Therefore, choose the value x @ -1.4 m.

9. (a) The spring constant of the diving board was
1.3 ´ 103 N/m.

(b) At this point, all of the elastic potential energy of the diving board is transferred to the diver as kinetic energy. The diver’s maximum speed will be 2.3 m/s.

(c) The diver’s kinetic energy is transformed into gravitational potential energy at the maximum height, which is 28 cm above the board.

10. (a) Maximum elastic potential energy of the spring:
29 J

CHAPTER 6 BLM ANSWER KEY

(b) Maximum velocity of the ball: approximately
5.4 m/s

(c) Maximum vertical height of the ball up the ramp: 1.5 m

11. (a)

12. P =

=

= / (1000 kg) (9.81 / m / ) (12 m)
s2
30 s

= 3900 W

13. efficiency = ´ 100%

Ein =

= 2500 J